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Let $V_i$ be a sequence of $k$ dimensional analytic subsets in $\mathbb C^N$. Suppose that the volume of $V_i$ is uniformly bounded, then Bishop's compactness theorem says that $V_i$ will convergence by sequence to an analytic subsets $V$.

Q1: What is the precise meaning of "converge" here. Q2: Is it possible that a sequence of singular point $q_i\in V_i$ converge to a smooth point $q\in V$. It seems impossible.

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Convergence is taken in Hausdorff sense, though you can define the structure of a complex variety (the Barlet space) on the set of cycles, taking every irreducible component with positive integer multiplicity. Barlet convergence is slightly more fine than the Hausdorff convergence, but not by much.

For the second, it is easy to construct an example of singular spaces converging to smooth. Take a sequence of curves in ${\Bbb C}P^2$ with each curve obtained as a union of two projective lines. Assume that this sequence converges to a union of a line with itself. Then the limit is smooth.

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  • $\begingroup$ Thanks @Misha, for the second question. Yes, you are right, probably, I think make it clear that non reduced means singular. So other counterexamples? $\endgroup$
    – xin fu
    Commented Jul 22, 2022 at 23:17
  • $\begingroup$ It all depends on your definitions, but if you want convergence in Barlet space and you consider all points in all cycles with multiplicities > 1 as "singular", then I suppose there should be such a result. I think that the Lelong number of a singular variety, considered as a current, is bigger in its singular point, and Lelong numbers are semi-continuous, so a limit of a singularity should be a singular point or a point with multiplicity > 1 (here I observe that convergence of currents is compatible with the Barlet space convergence). I never saw such a result, though. $\endgroup$
    – Misha Verbitsky
    Commented Jul 23, 2022 at 18:57

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